Integrate $\int x\sec^2(x)\tan(x)\,dx$ $$\int x\sec^2(x)\tan(x)\,dx$$
I just want to know what trigonometric function I need to use. I'm trying to integrate by parts.  My book says that the integral equals
$${x\over2\cos^2(x)}-{\sin(x)\over2\cos(x)}+C$$
 A: First you are going to need to use integration by parts... this will get the "x" out of the integral.
$\int x\sec^2 x\tan x \ dx$
$u = x, dv = \sec^2 x\tan x \ dx\\
du = dx, v = \frac 12 \sec^2 x$
How did I get v?
$v = \int \sec^2 x\tan x\ dx$
Now you will need to do a substitution $w = \tan x, dw = \sec^2 x$ or $w = \sec x, dw = \sec x\tan x$  either will work.
$\int x\sec^2 x\tan x \ dx\\
\frac 12 x\sec^2 x - \frac 12 \int \sec^2 x\ dx\\
\frac 12 x\sec^2 x - \frac 12 \tan x + c$
A: $$I=\int x\sec^2(x)\tan(x)dx$$
I will be walking you through this integral step-by-step. 
First, we integrate by parts: $$u=x\\du=dx$$
And $$dv=\sec^2(x)\tan(x)dx$$
Which is an integral I'd like to demonstrate 
$$v=\int dv\\v=\int\sec^2(x)\tan(x)dx$$
For this, we will use a $\omega$-substitution. Let $\omega=\tan(x)$. Therefore, $d\omega=\sec^2(x)dx$, giving
$$v=\int\omega d\omega=\frac{\omega^2}{2}\\v=\frac{\tan^2(x)}{2}$$
Next we finish our integration by parts:
$$I=uv-\int vdu\\I=\frac{x\tan^2(x)}{2}-\int\frac{\tan^2(x)}{2}dx$$
Then we define $A=\int\frac{\tan^2(x)}{2}dx$, which gives $I=\frac{x\tan^2(x)}{2}-A$.
Here we go with the next integral:
$$A=\int\frac{\tan^2(x)}{2}dx=\frac1{2}\int\tan^2(x)dx$$
Here we use a trig identity $\tan^2(x)=\sec^2(x)-1$.
$$A=\frac1{2}\int(\sec^2(x)-1)dx=\frac1{2}\int\sec^2(x)dx-\frac1{2}\int dx$$
$$A=\frac{\tan(x)-x}{2}$$
Thus, 
$$I=\frac{x\tan^2(x)-(\tan(x)-x)}2$$
$$I=\frac{x\tan^2(x)-\tan(x)+x}2$$
And like always, add your constant:
$$I=\frac{x\tan^2(x)-\tan(x)+x}2+C$$
A: Given $$\int x\sec^2(x)\tan(x)\ dx$$
Apply Integration By Parts $u=x$ and $v^{\prime}=\sec^2(x)\tan(x)$
and you get $$\dfrac12\tan^2(x)-\int\dfrac12\tan^2(x)\ dx$$
Can you take it from here?
A: Let $u = x, dv = \sec^2 x\tan xdx\implies v = \displaystyle \int\sec^2 x\tan xdx= \displaystyle \int \tan xd(\tan x)= \displaystyle \int wdw, w = \tan x$. Can you put it together ?
A: Hint: With $u = x$ and $dv = \sec^2 \tan(x)\,dx$, we have
$$
\int u \,dv = uv - \int v \,du = \frac 12 x \tan^2 x - \int \tan^2 x\,dx
$$
We can compute the integral of $\sec^2 \tan(x)$ with the $u$-substitution $u = \tan(x)$.
